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Deterministic finite automata of a language over alphabets {0, 1), which does not contain 3 consecutive 0’s. Minimum how many states, S, in all, the DFA will have and how many of them will be final states, F?
  • a)
    |S| = 5 and |F| = 2
  • b)
    |S| = 5 and |F| = 3
  • c)
    |S| = 4 and |F| = 3
  • d)
    |S| = 3 and |F|=1
Correct answer is option 'C'. Can you explain this answer?
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Deterministic finite automata of a language over alphabets {0, 1), whi...
The regular expresession for the language:
L = {W | W does not contain 3 consecutive 0 ’s} is 
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Deterministic finite automata of a language over alphabets {0, 1), whi...
To construct a deterministic finite automaton (DFA) that recognizes a language over the alphabet {0, 1} which does not contain three consecutive 0's, we can follow these steps:

1. Start by drawing a circle or a state for each possible state of the DFA. In this case, we can have three states:
- State 1: The current input does not contain any 0's.
- State 2: The current input contains one 0.
- State 3: The current input contains two consecutive 0's.

2. Label the initial state as the start state. In this case, State 1 is the start state.

3. Mark State 1 and State 2 as accepting states since the language allows any number of consecutive 1's.

4. Draw arrows from each state to the other states for each possible input symbol. In this case, we have two input symbols, 0 and 1.
- From State 1, draw an arrow labeled '0' to State 2. This transition represents encountering the first 0.
- From State 1, draw an arrow labeled '1' to State 1. This transition represents encountering any number of consecutive 1's.
- From State 2, draw an arrow labeled '0' to State 3. This transition represents encountering the second 0.
- From State 2, draw an arrow labeled '1' to State 1. This transition represents encountering any number of consecutive 1's.
- From State 3, draw an arrow labeled '0' to State 3. This transition represents encountering a third consecutive 0.
- From State 3, draw an arrow labeled '1' to State 1. This transition represents encountering any number of consecutive 1's.

5. The resulting DFA should look like this:

```
0 0
┌───────┐ ┌───────┐
│ │ │ │
──►│ State │──►│ State │
│ 1 │ │ 2 │
│ │ │ │
└───────┘ └───────┘
▲ │ │
│ │ │ 0
│ │ │
│ │ │
│ │ ▼
│ │ ┌─────────┐
│ │ │ │
└───┼►│ State │
1 │ │ 3 │
└───┼►│ │
1│ └─────────┘
│ ▲
│ │ 1
│ │
▼ ▼
```

This DFA will accept any string that does not contain three consecutive 0's.
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Deterministic finite automata of a language over alphabets {0, 1), which does not contain 3 consecutive 0’s. Minimum how many states, S, in all, the DFA will have and how many of them will be final states, F?a)|S| = 5 and |F| = 2b)|S| = 5 and |F| = 3c)|S| = 4 and |F| = 3d)|S| =3 and |F|=1Correct answer is option 'C'. Can you explain this answer?
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Deterministic finite automata of a language over alphabets {0, 1), which does not contain 3 consecutive 0’s. Minimum how many states, S, in all, the DFA will have and how many of them will be final states, F?a)|S| = 5 and |F| = 2b)|S| = 5 and |F| = 3c)|S| = 4 and |F| = 3d)|S| =3 and |F|=1Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Deterministic finite automata of a language over alphabets {0, 1), which does not contain 3 consecutive 0’s. Minimum how many states, S, in all, the DFA will have and how many of them will be final states, F?a)|S| = 5 and |F| = 2b)|S| = 5 and |F| = 3c)|S| = 4 and |F| = 3d)|S| =3 and |F|=1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Deterministic finite automata of a language over alphabets {0, 1), which does not contain 3 consecutive 0’s. Minimum how many states, S, in all, the DFA will have and how many of them will be final states, F?a)|S| = 5 and |F| = 2b)|S| = 5 and |F| = 3c)|S| = 4 and |F| = 3d)|S| =3 and |F|=1Correct answer is option 'C'. Can you explain this answer?.
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